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24 Dec 2025, 04:52
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Statement 1
yes: 1,2,3,4,5,6,7 = 28
No: 0,1,2,3,4,5,6,13 = 28
Both outcomes possible
Statement 1 not sufficient
Statement 2
Allows yes: 7,5,5,4,4,2,1= all > or = 1
No: 7,7,7,7,0,0,0= some agents get 0
Not sufficient
Statement 1&2
7 Agents
7 distinct number
maximum=7
Total=28
Oly possible distinct set of 7 non negative integers with maximum7 tat sums to 28 is:
1,2,3,4,5,6,7
so every agent gets at least 1 ticket
Answer: Statement 1&2 together are sufficient but neither alone is sufficient
yes: 1,2,3,4,5,6,7 = 28
No: 0,1,2,3,4,5,6,13 = 28
Both outcomes possible
Statement 1 not sufficient
Statement 2
Allows yes: 7,5,5,4,4,2,1= all > or = 1
No: 7,7,7,7,0,0,0= some agents get 0
Not sufficient
Statement 1&2
7 Agents
7 distinct number
maximum=7
Total=28
Oly possible distinct set of 7 non negative integers with maximum7 tat sums to 28 is:
1,2,3,4,5,6,7
so every agent gets at least 1 ticket
Answer: Statement 1&2 together are sufficient but neither alone is sufficient
24 Dec 2025, 05:39
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did each agent get a ticket?
7 agents, 28 tickets to each agent.
Testing statement 1
Maybe NO 0,1,2,3,4,8,10 sum =28 including 0
Maybe YES 1,2,3,4,5,6,7 sum =28 >1
NOT SUFFICIENT
Statement 2
Maybe NO 7,7,7,7,0,0,0 sum 28 some got 0
Maybe YES. 7,7,5,4,3,1,1 sum=28 all >1
INSUFFICIENT
both
count max of 7 choose integers from(0,1,2,3,4,5,6,7)
to get to 28 with 7 numbers and max of 7 (1,2,3,4,5,6)
including 0 omitting k 28-k<28
each agent received at least 1 ticket
SUFFICIENT
C. both statements together are sufficient, but each alone is not.
7 agents, 28 tickets to each agent.
Testing statement 1
Maybe NO 0,1,2,3,4,8,10 sum =28 including 0
Maybe YES 1,2,3,4,5,6,7 sum =28 >1
NOT SUFFICIENT
Statement 2
Maybe NO 7,7,7,7,0,0,0 sum 28 some got 0
Maybe YES. 7,7,5,4,3,1,1 sum=28 all >1
INSUFFICIENT
both
count max of 7 choose integers from(0,1,2,3,4,5,6,7)
to get to 28 with 7 numbers and max of 7 (1,2,3,4,5,6)
including 0 omitting k 28-k<28
each agent received at least 1 ticket
SUFFICIENT
C. both statements together are sufficient, but each alone is not.
24 Dec 2025, 05:59
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Except for 28 tickets per 7 agents, we don't have much to take from the task itself.
(1) There's only 1 arrangement when everyone gets at least 1 ticket: \(1+2+3+4+5+6+7=28.\)
However, if someone gets 0, we see \(0+1+2+3+4+5+6+x=28,\) where \(x=7\), and these seven tickets can be distributed among the agents in many ways to keep the numbers unique. Therefore, it's insfficient.
(2) We know that someone got 7. Well, if someone else got 6, and three more got 5, then 2 agents could easily stay without work. At the same time, we could spread the tickets 1-through-7 among each of them. It's clearly insufficient.
(1) + (2) Combining the two options, we must land at the only arrangement: \(1+2+3+4+5+6+7=28\), which means everyone had some work. Sufficient, so the answer is C.
(1) There's only 1 arrangement when everyone gets at least 1 ticket: \(1+2+3+4+5+6+7=28.\)
However, if someone gets 0, we see \(0+1+2+3+4+5+6+x=28,\) where \(x=7\), and these seven tickets can be distributed among the agents in many ways to keep the numbers unique. Therefore, it's insfficient.
(2) We know that someone got 7. Well, if someone else got 6, and three more got 5, then 2 agents could easily stay without work. At the same time, we could spread the tickets 1-through-7 among each of them. It's clearly insufficient.
(1) + (2) Combining the two options, we must land at the only arrangement: \(1+2+3+4+5+6+7=28\), which means everyone had some work. Sufficient, so the answer is C.
